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\begin{document}
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\vskip 1cm{\LARGE\bf
The Minimal Density of a Letter in an \\
\vskip .1in
Infinite Ternary Square-Free Word
is $\frac{883}{3215}$
}
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\large
Andrey Khalyavin \\
Mech. \& Math. Department\\
Moscow State University\\
119992 Moscow \\
Russia\\
\href{mailto:halyavin@land.ru}{\tt halyavin@land.ru} \\
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\begin{abstract}
The problem of determining the minimal density of a letter in an infinite
ternary square-free word was investigated by
Tarannikov and Ochem.
In this paper we solve this problem, and prove
that the minimal density is equal to $\frac{883}{3215}$.
\end{abstract}
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\section{Introduction}
A word is called {\it square-free} if it cannot be written in the form $axxb$
for two words $a$, $b$ and a nonempty word $x$. In this paper we study the
minimal density of a letter $a$ in an infinite square-free word over the
alphabet $\{a,b,c\}$. The {\it density} of a letter $a$ in a finite word $w$ is
$\frac{n(w)}{|w|}$ where $n(w)$ is the number of letters $a$ in $w$ and
$|w|$ is the length of the word. The {\it density} of a letter $a$ in an
infinite word is $\liminf\limits_{l\rightarrow\infty}\frac{n(w_l)}{|w_l|}$
where $w_l$ is the prefix of the word $w$ of length $l$. Tarannikov
\cite{taran} found the minimal density of a letter $a$ with up to $4$ tight
digits after the decimal point. Ochem \cite{ochem} improved this result,
proving that the minimal density of a letter $a$ belongs to the interval
$[\frac{1000}{3641},\frac{883}{3215}]$. In this article we completely solve
this problem and prove that the minimal density of a letter $a$ is equal to
$\frac{883}{3215}$. We use huge computer calculations in our proof. You can
find sources of the programs at
{\tt http://mech.math.msu.su/department/dm/dmmc/PUBL/words.zip}.
\section{Main result}
Our aim is to prove that the minimal density is at least
$\rho=\frac{883}{3215}$. Let us construct $4$ special words $b_1,\dots,b_4$ of
length $3215$ that contain $883$ letters $a$ each (they are probably the same
words used by Ochem \cite{ochem} in order to construct an infinite word
with density $\rho$). We call these words the {\it basic words}.The files
with these basic words can be found at the URL address above. We proved the
next theorem by a computation using the technique of backtracking.
\begin{theorem}
In any square-free word of length 90000 there exist either
\begin{itemize}\item[1)]a prefix with density greater than or equal
than $\rho$,
\item[2)]two adjacent basic words (a subword $b_ib_j$).
\end{itemize}
\label{sqfree1}
\end{theorem}
So, either an infinite square-free word can be split into words with densities
greater than or equal than $\rho$ or there exist two basic words one after another. In the
first case the density of an infinite word is greater than or equal than $\rho$ because the
lengths of the words in the splitting are bounded by 90000. So we can assume
that an infinite square-free word begins with a word $b_ib_j$.
When our program that proves Theorem \ref{sqfree1} finds two adjacent basic
words in the current word $w$, it saves the pair $(n(w),|w|)$ and backtracks.
Let us denote by $W$ the set of all such current words. Then we delete equal
pairs $(n(w),|w|)$ from the list and obtain $80$ distinct pairs $(n(w),|w|)$.
Let us denote by $L$ the set of these pairs. Then for all $l<200000$ we
calculate the maximal number $p(l)$ such that after the concatenation of any
words $w\in W$ and $v$ such that $n(v)\ge p(l),|v|=l$, the word $vw$ has the
density greater than or equal than $\rho$. So,
$p(l)=\max_{(n(w),|w|)\in L}[\rho(l+|w|)]-n(w)+1$. Our second program proves
the next theorem.
\begin{theorem}
Let $u$ be an infinite square-free word that begins with a word
$b_ib_j$. Then $u$ contains one of the following prefixes:
\begin{itemize}\item[1)] a prefix $w$ such that $2\cdot 3215